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Apr 20, 2023 at 16:24 comment added Student Thanks a lot for your answer, this has been very helpful!
Apr 20, 2023 at 16:20 vote accept Student
Apr 20, 2023 at 13:54 comment added Willie Wong For your question that is completely unimportant. The proof that @MaoWao provided works equally well on compact manifolds, possibly with positive curvature. But on compacts since the spectrum is discrete you will have spectral gap and the convergence is exponential.
Apr 20, 2023 at 13:50 comment added Overflowian By the way, you didn't use the extra hypothesis of non-positive curvature, didn't you?
Apr 20, 2023 at 12:55 comment added MaoWao Yes, essential self-adjointness is stable under additive bounded (self-adjoint) perturbations.
Apr 20, 2023 at 12:28 comment added Overflowian So $-\Delta-\lambda_1$ is positive semi-definite, does it still have a unique extension?
Apr 20, 2023 at 12:24 comment added MaoWao I took $\lambda_1$ to mean the bottom of the spectrum of $-\Delta$ (which is actually contained in the spectrum because it is closed). Of course, if $\lambda_1$ is embedded in the spectrum, this is no longer true -- then you can have approximate eigenfunctions of the bottom of the spectrum for which $e^{t(\Delta+\lambda_1)}u_0$ blows up in $L^2$ norm.
Apr 20, 2023 at 12:05 comment added Overflowian I'm a bit confused by signs here, you seems to say that the spectrum of $-\Delta $ satisfies $\sigma(-\Delta)\subset (\lambda_1,+\infty)$. So is the definition of $\lambda_1 $ given by $\inf \sigma(-\Delta)$?. I thought that it was the smallest positive eigenvalue and in particular it should be contained in the spectrum.
Apr 20, 2023 at 9:43 history edited MaoWao CC BY-SA 4.0
added 68 characters in body
Apr 20, 2023 at 6:38 history answered MaoWao CC BY-SA 4.0